Fall 2005
Lecture 1: Introduction
Intro. to Computer Language Engineering
Course Administration info.
Outline
• Course Administration Information
• Introduction to computer language engineering
– What are compilers?
– Why should we learn about them?
– Anatomy of a compiler
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Course Administration
•
•
•
•
•
•
Staff
Text
Course Outline
The Project
Project Groups
Grading
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Staff
• Lecturers
– Prof. Saman Amarasinghe
– Prof. Martin Rinard
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Optional Textbooks
Tiger • Modern Compiler Implementation in Java
by A.W. Appel
Book
Cambridge University Press, 1998
Whale • Advanced Compiler Design and Implementation
by Steven Muchnick
Book
Morgan Kaufman Publishers, 1997
Dragon
Book
• Compilers -- Principles, Techniques and Tools
by Aho, Sethi and Ullman
Addison-Wesley, 1988
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The Five Segments
n
Lexical and Syntax Analysis
o
Semantic Analysis
p
Code Generation
q
Data-flow Optimizations
r
Instruction Optimizations
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Each Segment...
• Segment Start
– Project Description
• Lectures
– 2 to 4 lectures
•
•
•
•
Project Time
(Project checkpoint)
Project Time
Project Due
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Project Groups
• Each project group consists of 3 to 4 students
• Grading
– All group members (mostly) get the same grade
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Weekly Group Meeting with the TA
• TA will schedule a weekly 30 minute slot
• The group will get to:
–
–
–
–
Ask questions about the project
Discuss design decisions
Describe what each person is doing
Answer questions the TA has about the project
• TA will use this to measure individual
contribution to the project Æ be there!
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Grades
• Compiler project
58%
• Paper Discussion
12% (3% each)
• In-class Quizzes
30% (10% each)
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Grades for the Project
–
–
–
–
–
Scanner/Parser
Semantic Checking
Code Generation
Data-flow Opt.
Instruction Opt.
Saman Amarasinghe
10%
10%
14%
12%
12%
58%
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The Quiz
• Three Quizzes
• In-Class Quiz
– 50 Minutes (be on time!)
– Open book, open notes
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Paper Discussions
• Will be giving out 3 papers
• Read the paper, think about…
–
–
–
–
–
–
–
Do you like the research?
What is the context of the research?
Did anything surprise you about the paper?
Do the results allow you to evaluate the proposed technique?
Do the authors understand the wider ramifications of their research?
How have things changed since the paper was written?
How important will the research be in the future?
• Write a 150 to 200 word summary of the paper.
• Will schedule a one-on-one (15 or 20 minute) meeting with a
professor or a TA to talk about the paper.
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Outline
• Course Administration Information
• Introduction to computer language engineering
– What are compilers?
– Why should we learn about them?
– Anatomy of a compiler
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What is Computer Language
Engineering
1. How to give instructions to a computer
2. How to make the computer carryout the
instructions efficiently
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Power of a Language
• Can use to describe any action
– Not tied to a “context”
• Many ways to describe the same action
– Flexible
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How to instruct a computer
• How about natural languages?
–
–
–
–
English??
“Open the pod bay doors, Hal.”
“I am sorry Dave, I am afraid I cannot do that”
We are not there yet!!
• Natural Languages:
– Same expression describes many possible actions
– Ambiguous
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How to instruct a computer
• How about natural languages?
–
–
–
–
English??
“Open the pod bay doors, Hal.”
“I am sorry Dave, I am afraid I cannot do that”
We are not there yet!!
• Use a programming language
– Examples: Java, C, C++, Pascal, BASIC, Scheme
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Programming Languages
• Properties
–
–
–
–
need to be precise
need to be concise
need to be expressive
need to be at a high-level (lot of abstractions)
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High-level Abstract Description to Low-level Implementation Details
My poll ratings are low,
lets invade a small nation
President
Cross the river and take
defensive positions
General
Forward march, turn left
Stop!, Shoot
Sergeant
Foot Soldier
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Figures by MIT OCW.
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©MIT Fall 1998
1. How to instruct the computer
• Write a program using a programming language
– High-level Abstract Description
• Microprocessors talk in assembly language
– Low-level Implementation Details
Program
written
in a
Programming
Languages
Saman Amarasinghe
Compiler
21
Assembly
Language
Translation
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1. How to instruct the computer
• Input: High-level programming language
• Output: Low-level assembly instructions
• Compiler has to:
–
–
–
–
Read and understand the program
Precisely determine what actions it require
Figure-out how to carry-out those actions
Instruct the computer to carry out those actions
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Example (input program)
int expr(int n)
{
int d;
d = 4 * n * n * (n + 1) * (n + 1);
return d;
}
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Example (Output assembly code)
lda $30,-32($30)
stq $26,0($30)
stq $15,8($30)
bis $30,$30,$15
bis $16,$16,$1
stl $1,16($15)
lds $f1,16($15)
sts $f1,24($15)
ldl $5,24($15)
bis $5,$5,$2
s4addq $2,0,$3
ldl $4,16($15)
mull $4,$3,$2
ldl $3,16($15)
addq $3,1,$4
mull $2,$4,$2
ldl $3,16($15)
addq $3,1,$4
mull $2,$4,$2
stl $2,20($15)
ldl $0,20($15)
br $31,$33
$33:
bis $15,$15,$30
ldq $26,0($30)
ldq $15,8($30)
addq $30,32,$30
ret $31,($26),1
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Efficient Execution of the Actions
Cross the river and take
defensive positions
General
Sergeant
Foot Soldier
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Figures by MIT OCW.
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©MIT Fall 1998
Efficient Execution of the Actions
Cross the river and take
defensive positions
General
Where to cross the river? Use the
bridge upstream or surprise the enemy
by crossing downstream?
How do I minimize the casualties??
Sergeant
Foot Soldier
Figures by MIT OCW.
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Efficient Execution of the Actions
My poll ratings are low,
lets invade a small nation
President
Russia or Bermuda?
Or just stall for his poll
numbers to go up?
General
Figures by MIT OCW.
Saman Amarasinghe
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Efficient Execution of the Actions
• Mapping from High to Low
– Simple mapping of a program to assembly language
produces inefficient execution
– Higher the level of abstraction ⇒ more inefficiency
• If not efficient
– High-level abstractions are useless
• Need to:
– provide a high level abstraction
– with performance of giving low-level instructions
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Example
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
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test:
subu $fp, 16
sw
zero, 0($fp)
sw
zero, 4($fp)
sw
zero, 8($fp)
lab1:
mul
$t0, $a0, 4
div
$t1, $t0, $a1
lw
$t2, 8($fp)
mul
$t3, $t1, $t2
lw
$t4, 8($fp)
addui$t4, $t4, 1
lw
$t5, 8($fp)
addui$t5, $t5, 1
mul
$t6, $t4, $t5
addu $t7, $t3, $t6
lw
$t8, 0($fp)
add $t8, $t7, $t8
sw
$t8, 0($fp)
lw
$t0, 4($fp)
mul
$t1, $t0, a1
lw
$t2, 0($fp)
add
$t2, $t2, $t1
sw
$t2, 0($fp)
lw
$t0, 8($fp)
addui $t0, $t0, 1
sw
$t0, 8($fp)
ble
$t0, $a3, lab1
lw
$v0, 0($fp)
addu
$fp, 16
b
$ra
Saman Amarasinghe
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#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
x = 0
y = 0
i = 0
for(i=0;i<N; i++)
a*4
a*4/b
i
a*4/b*i
i
i+1
i
i+1
(i+1)*(i+1)
a*4/b*i + (i+1)*(i+1)
x
x = x + a*4/b*i + (i+1)*(i+1)
#
#
#
#
y
b*y
x
x = x + b*y
# i
# i+1
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Lets Optimize...
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
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Constant Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
Saman Amarasinghe
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Constant Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
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Constant Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
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Constant Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*y;
}
return x;
}
Saman Amarasinghe
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Constant Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*0;
}
return x;
}
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*0;
}
return x;
}
Saman Amarasinghe
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + b*0;
}
return x;
}
Saman Amarasinghe
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + 0;
}
return x;
}
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + 0;
}
return x;
}
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x + 0;
}
return x;
}
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Algebraic Simplification
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x;
}
return x;
}
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Copy Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
x = x;
}
return x;
}
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Copy Propagation
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
}
return x;
}
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Common Subexpression Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
}
return x;
}
Saman Amarasinghe
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©MIT Fall 1998
Common Subexpression Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x + (4*a/b)*i + (i+1)*(i+1);
}
return x;
}
Saman Amarasinghe
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©MIT Fall 1998
Common Subexpression Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y, t;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + (i+1)*(i+1);
}
return x;
}
Saman Amarasinghe
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©MIT Fall 1998
Common Subexpression Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y, t;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
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Dead Code Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y, t;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
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Dead Code Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y, t;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
Saman Amarasinghe
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©MIT Fall 1998
Dead Code Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, y, t;
x = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
Saman Amarasinghe
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Dead Code Elimination
int sumcalc(int a, int b, int N)
{
int i;
int x, t;
x = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
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Loop Invariant Removal
int sumcalc(int a, int b, int N)
{
int i;
int x, t;
x = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
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©MIT Fall 1998
Loop Invariant Removal
int sumcalc(int a, int b, int N)
{
int i;
int x, t;
x = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + (4*a/b)*i + t*t;
}
return x;
}
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©MIT Fall 1998
Loop Invariant Removal
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u;
x = 0;
u = (4*a/b);
for(i = 0; i <= N; i++) {
t = i+1;
x = x + u*i + t*t;
}
return x;
}
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©MIT Fall 1998
Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u;
x = 0;
u = (4*a/b);
for(i = 0; i <= N; i++) {
t = i+1;
x = x + u*i + t*t;
}
return x;
}
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©MIT Fall 1998
Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u;
u*0,
x = 0;
u*1,
u = (4*a/b);
u*2,
for(i = 0; i <= N; i++) {
u*3,
t = i+1;
u*4,
x = x + u*i + t*t;
...
}
return x;
}
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v=0,
v=v+u,
v=v+u,
v=v+u,
v=v+u,
...
©MIT Fall 1998
Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = (4*a/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + u*i + t*t;
v = v + u;
}
return x;
}
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Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = (4*a/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
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Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = (4*a/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
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Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = (4*a/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
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©MIT Fall 1998
Strength Reduction
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = ((a<<2)/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
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©MIT Fall 1998
Register Allocation
fp
Local variable X
Local variable Y
Local variable I
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Register Allocation
fp
Local variable X
Local variable Y
Local variable I
$t9
$t8
$t7
$t6
$t5
=
=
=
=
=
X
t
u
v
i
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Optimized Example
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = ((a<<2)/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
Saman Amarasinghe
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6.035
©MIT Fall 1998
test:
subu
add
sll
div
add
add
$fp,
$t9,
$t0,
$t7,
$t6,
$t5,
lab1:
addui$t8,
mul $t0,
addu $t1,
addu $t9,
16
zero, zero
$a0, 2
$t0, $a1
zero, zero
zero, zero
#
#
#
#
#
x = 0
a<<2
u = (a<<2)/b
v = 0
i = 0
$t5, 1
$t8, $t8
$t0, $t6
t9, $t1
#
#
#
#
#
for(i=0;i<N; i++)
t = i+1
t*t
v + t*t
x = x + v + t*t
addu $6, $6, $7
# v = v + u
addui$t5, $t5, 1
ble $t5, $a3, lab1
# i = i+1
addu $v0, $t9, zero
addu $fp, 16
b
$ra
Saman Amarasinghe
66
6.035
©MIT Fall 1998
Unoptimized Code
Optimized Code
test:
test:
subu
sw
sw
sw
$fp, 16
zero, 0($fp)
zero, 4($fp)
zero, 8($fp)
mul
div
lw
mul
lw
addui
lw
addui
mul
addu
lw
add
sw
lw
mul
lw
add
sw
lw
addui
sw
ble
$t0,
$t1,
$t2,
$t3,
$t4,
$t4,
$t5,
$t5,
$t6,
$t7,
$t8,
$t8,
$t8,
$t0,
$t1,
$t2,
$t2,
$t2,
$t0,
$t0,
$t0,
$t0,
lw
addu
b
$v0, 0($fp)
$fp, 16
$ra
lab1:
$a0, 4
$t0, $a1
8($fp)
$t1, $t2
8($fp)
$t4, 1
8($fp)
$t5, 1
$t4, $t5
$t3, $t6
0($fp)
$t7, $t8
0($fp)
4($fp)
$t0, a1
0($fp)
$t2, $t1
0($fp)
8($fp)
$t0, 1
8($fp)
$a3, lab1
$fp,
$t9,
$t0,
$t7,
$t6,
$t5,
16
zero, zero
$a0, 2
$t0, $a1
zero, zero
zero, zero
addui
mul
addu
addu
addu
addui
ble
$t8, $t5, 1
$t0, $t8, $t8
$t1, $t0, $t6
$t9, t9, $t1
$6, $6, $7
$t5, $t5, 1
$t5, $a3, lab1
addu
addu
b
$v0, $t9, zero
$fp, 16
$ra
lab1:
4*ld/st + 2*add/sub + br +
N*(9*ld/st + 6*add/sub + 4* mul + div + br)
= 7 + N*21
Execution time = 43 sec
Saman Amarasinghe
subu
add
sll
div
add
add
6*add/sub + shift + div + br +
N*(5*add/sub + mul + br)
= 9 + N*7
Execution time = 17 sec
67
6.035
©MIT Fall 1998
Compilers Optimize Programs for…
•
•
•
•
•
•
Performance/Speed
Code Size
Power Consumption
Fast/Efficient Compilation
Security/Reliability
Debugging
Saman Amarasinghe
68
6.035
©MIT Fall 1998
Compilers help increase the
level of abstraction
• Programming languages
– From C to OO-languages
with garbage collection
– Even more abstract
definitions
• Microprocessor
– From simple CISC to RISC
to VLIW to ….
Saman Amarasinghe
69
6.035
©MIT Fall 1998
Anatomy of a Computer
Program
written
in a
Programming
Languages
Saman Amarasinghe
Compiler
70
Assembly
Language
Translation
6.035
©MIT Fall 1998
Anatomy of a Computer
Program (character stream)
Lexical Analyzer (Scanner)
Token Stream
Saman Amarasinghe
71
6.035
©MIT Fall 1998
Lexical Analyzer (Scanner)
2 3 4
*
( 1 1
+ - 2 2 )
Num(234) mul_op lpar_op Num(11) add_op Num(-22) rpar_op
18..23 + val#ue
Saman Amarasinghe
72
6.035
©MIT Fall 1998
Lexical Analyzer (Scanner)
2 3 4
*
( 1 1
+ - 2 2 )
Num(234) mul_op lpar_op Num(11) add_op Num(-22) rpar_op
18..23 + val#ue
Variable names cannot have ‘#’ character
Not a number
Saman Amarasinghe
73
6.035
©MIT Fall 1998
Anatomy of a Computer
Program (character stream)
Lexical Analyzer (Scanner)
Token Stream
Syntax Analyzer (Parser)
Parse Tree
Saman Amarasinghe
74
6.035
©MIT Fall 1998
Syntax Analyzer (Parser)
num ‘*’ ‘(‘ num ‘+’ num ‘)’
<expr>
<expr>
num
<op>
*
<expr>
<expr>
(
<expr> <op>
num
Saman Amarasinghe
75
+
)
<expr>
num
6.035
©MIT Fall 1998
Syntax Analyzer (Parser)
int * foo(i, j, k))
int i;
int j;
Extra parentheses
{
for(i=0; i j) {
fi(i>j)
Missing increment
return j;
Not an expression
}
Not a keyword
Saman Amarasinghe
76
6.035
©MIT Fall 1998
Anatomy of a Computer
Program (character stream)
Lexical Analyzer (Scanner)
Token Stream
Syntax Analyzer (Parser)
Parse Tree
Semantic Analyzer
Intermediate Representation
Saman Amarasinghe
77
6.035
©MIT Fall 1998
Semantic Analyzer
int * foo(i, j, k)
int i;
int j;
{
int x;
x = x + j + N;
return j;
}
Saman Amarasinghe
78
Type not declared
Mismatched return type
Uninitialized variable used
Undeclared variable
6.035
©MIT Fall 1998
Anatomy of a Computer
Program (character stream)
Lexical Analyzer (Scanner)
Token Stream
Syntax Analyzer (Parser)
Parse Tree
Semantic Analyzer
Intermediate Representation
Code Optimizer
Optimized Intermediate Representation
Saman Amarasinghe
79
6.035
©MIT Fall 1998
Optimizer
int sumcalc(int a, int b, int N)
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = ((a<<2)/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
{
int i;
int x, y;
x = 0;
y = 0;
for(i = 0; i <= N; i++) {
x = x+4*a/b*i+(i+1)*(i+1);
x = x + b*y;
}
return x;
}
Saman Amarasinghe
80
6.035
©MIT Fall 1998
Anatomy of a Computer
Program (character stream)
Lexical Analyzer (Scanner)
Token Stream
Syntax Analyzer (Parser)
Parse Tree
Semantic Analyzer
Intermediate Representation
Code Optimizer
Optimized Intermediate Representation
Code Generator
Assembly code
Saman Amarasinghe
81
6.035
©MIT Fall 1998
Code Generator
int sumcalc(int a, int b, int N)
{
int i;
int x, t, u, v;
x = 0;
u = ((a<<2)/b);
v = 0;
for(i = 0; i <= N; i++) {
t = i+1;
x = x + v + t*t;
v = v + u;
}
return x;
}
Saman Amarasinghe
82
test:
subu
add
sll
div
add
add
lab1:
addui
$fp,
$t9,
$t0,
$t7,
$t6,
$t5,
16
zero, zero
$a0, 2
$t0, $a1
zero, zero
zero, zero
$t8, $t5, 1
mul
addu
addu
$t0, $t8, $t8
$t1, $t0, $t6
$t9, t9, $t1
addu
$6, $6, $7
addui
$t5, $t5, 1
ble
$t5, $a3, lab1
addu
addu
b
$v0, $t9, zero
$fp, 16
$ra
6.035
©MIT Fall 1998